# Maximum of norm inducted by inner product

I want to choose such two norms $$\Vert\cdot\Vert_a, \; \Vert\cdot\Vert_b$$ for which:

$$\Vert\cdot\Vert_c := \max\{\Vert\cdot\Vert_a, \; \Vert\cdot\Vert_b\}$$

Is a norm inducted by inner product.

My work so far

According to Maximum of two norms is norm I'm asured that my object is well defined. Norm is inducted by inner product when the following formula is satisfied:

$$\Vert x+y \Vert^2 + \Vert x-y \Vert^2 = 2(\Vert x \Vert^2 + \Vert y \Vert^2)$$

Let's define $$A := (x_1, y_1)$$, $$B:= (x_2, y_2)$$ and rewrite the formula above:

$$\Vert A+B\Vert_c^2 + \Vert A-B \Vert_c^2 = 2(\Vert A\Vert_c^2 + \Vert B \Vert_c^2)$$

$$\Vert (x_1+x_2, y_1+y_2)\Vert_c^2 + \Vert (x_1-x_2, y_1-y_2)\Vert_c^2 = 2(\Vert (x_1, y_1)\Vert_c^2+\Vert (x_2, y_2)\Vert_c^2)$$

$$\max{\{\Vert x_1 + x_2 \Vert_a, \Vert y_1 + y_2 \Vert_b\}}^2 + \max{\{\Vert x_1 - x_2 \Vert_a, \Vert y_1 - y_2 \Vert_b\}}^2 = 2( \max{\{\Vert x_1 \Vert_a, \Vert y_1 \Vert_b\}}^2 + \max{\{\Vert x_2 \Vert_a, \Vert y_2 \Vert_b\}}^2)$$

And here I wasnt sure which norms $$a, b$$ to chose. I tried some combinations $$\mathbb{l}_1$$ with supremum but it doesn't give me more than just bizzare looking expressions.

Could you please help me findin some norms $$a$$ and $$b$$ for which parallelogram law holds ?

Here is a different idea. The unit ball of the $$\|\cdot\|_c$$ norm is equal to the intersection of the unit balls of the $$\|\cdot\|_a$$ and $$\|\cdot\|_b$$ norms. Now choose two norms such that the intersection of their unit balls is the Euclidean unit ball.

Here is one on $$\mathbb R^2$$: $$\|(x_1,x_2)\|_a:= \begin{cases} \sqrt{x_1^2 + x_2^2} & \text{if }x_1x_2 \ge 0\\\max(|x_1|,|x_2|) & \text{if }x_1x_2 < 0. \end{cases}$$ Define $$\|(x_1,x_2)\|_b:=\|(-x_1,x_2)\|_a$$. Since $$\max(|x_1|,|x_2|)\le\sqrt{x_1^2 + x_2^2}$$, we have $$\max(\|(x_1,x_2)\|_a,\|(x_1,x_2)\|_b) = \sqrt{x_1^2+x_2^2}$$.

• Hey @daw thank you very much for your answer! I just have one question related to it. How do you know that $\max(\|(x_1,x_2)\|_a,\|(x_1,x_2)\|_b) = \sqrt{x_1^2+x_2^2}$ ? i.e. it's just assuming that $x_1x_2 \ge 0$. Besides you defined norm $b$ as let's use word 'reversed' norm $a$ i.e. $\|(x_1,x_2)\|_b:=\|(x_2,x_1)\|_a$. However it doesn't make any difference right ? Due to the fact that $\|(x_2,x_1)\|_a = \|(x_1,x_2)\|_a$ Mar 23 '21 at 17:12
• I understand you're saying that for such defined norms $\Vert \cdot \Vert_a$, $\Vert \cdot \Vert_b$ one of them will be $\sqrt{x_1^2 + x_2^2}$ and second one $\max(|x_1|, |x_2|)$. Due to fact that $\sqrt{x_1^2 + x_2^2} \ge \max(|x_1|, |x_2|)$ - maximum of those (so norm $\Vert \cdot \Vert_c$) will be always $\sqrt{x_1^2 + x_2^2}$. The only thing which is not make fully sense for me it's that one of those norms will be in format $\sqrt{x_1^2 + x_2^2}$ and second one in maximum. In my opinion norms $a$ and $b$ are equal. Mar 23 '21 at 18:44
• just, you are right. Have to flip the sign of one coordinate for the $b$-norm.
– daw
Mar 23 '21 at 18:58
• Perfect! Now it makes sense! Maybe the last question to be as formal as possible. How do you know that such defined norm $\Vert \cdot \Vert_a$ is a norm ? Mar 23 '21 at 20:09
• :) Tried triangle inequality on paper. At the end only one case is important, should work with some case distinctions on signs.
– daw
Mar 23 '21 at 20:18